Visible light of wavelength 6000 $\times$ 10-8 cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at 60o from the central maximum. If the first minimum is produced at $\theta$1, then $\theta$1, is close to :
Solution
For 2<sup>nd</sup> minima
<br><br>$\sin \theta = {{2\lambda } \over d }$
<br><br>$\Rightarrow$ $\sin 60^\circ = {{2\lambda } \over d}$
<br><br>$\Rightarrow$ ${\lambda \over d} = {{\sqrt 3 } \over 4}$
<br><br>For 1<sup>st</sup> minima
<br><br>$\sin {\theta _1} = {\lambda \over d}$ = ${{\sqrt 3 } \over 4}$
<br><br>$\Rightarrow$ ${\theta _1} =$ 25<sup>o</sup>
About this question
Subject: Physics · Chapter: Optics · Topic: Reflection and Mirrors
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